question_answer

Construct a \[\Delta \]PQR, where PQ = 5 cm, \[\angle \]P =\[~45{}^\circ \] and \[\angle \]Q = \[75{}^\circ \]

Answer:

Steps of Construction:

(a) Draw a line segment PQ = 5 cm.

(b) At P, construct\[\angle QPX=45{}^\circ \].

(c) At Q construct\[\angle PQY=75{}^\circ \].

(d) Let the rays \[\overline{PX}\]and \[\overline{QY}\]intersect at R.

Thus \[\Delta PQR\]is the required triangle.